# Stuck on Thevenin Equivalence Problem [closed]

I'm supposed to find the necessary elements for the Thevenin equivalent of the pictured circuit. That's $$\V_{TH}\$$, $$\R_{TH}\$$, and the Norton current ($$\I_N\$$). I'm also required to show that $$\\frac{V_{TH}}{R_{TH}} = I_N\$$. This is where I run into problems. From what I've been able to find, $$\V_{TH} = I R_2\$$, $$\R_{TH} = R_1\$$, and $$\I_N = \frac{R_2}{R_1 + R_2}\$$. This clearly doesn't obey the equation above. However, I haven't been able to find any other results for each variable, and they all seem to match up with simulations of the circuit. I've been stuck here for a while, please let me know if I'm missing something.

• Check your $R_{TH}$ and $I_N$. Commented Sep 5 at 20:41

Your $$\R_{TH}\$$ seems to be incorrect. When we remove the current source (replacing it by an open circuit), we essentially get between $$\A\$$ and $$\B\$$ two resistors in series. In particular, we have $$\R_{TH} = R_1 \color{red}{+ R_2}\$$.

Then if we want to find $$\I_N\$$ independently, we find the short-circuit current for terminal $$\A - B\$$. The circuit is just a current divider. We get $$\I_N = \frac{R_2}{R_1 + R_2} \color{red}{I}\$$.

Now, we can verify that

$$\frac{V_{TH}}{R_{TH}} = \frac{I R_2}{R_1 + R_2} = \frac{R_2}{R_1 + R_2} I = I_N$$

as desired.

• Okay, I think I see where I differed, and it has me somewhat confused. You say to to open the circuit when removing the current source, which of course will lead to R_Th = R1+R2. My professor has had us replace sources with a short when finding R_Th, which will lead to there being no "current" through R2 when measuring resistance across the ports. Is this different when working with current sources? Commented Sep 5 at 23:22
• @ModestasBotha Yeah, when eliminating the current source, you must open the circuit. Current sources have infinite impedance, whereas voltage sources have zero impedance. Commented Sep 6 at 0:48

To find the thevenin equivalent resistance, open circuit the current source.short circuit any voltage sources while keeping dependent current and voltage sources as they are. In the circuit above, the thevenin resistance is $$\R_{Th} = R_1 + R_2\$$.

The Thevenin voltage across a-b is $$\IR_2\$$.

You can find $$\\frac{V_{Th}}{R_{Th}}\$$ from above.

From what I've been able to find, $$\V_{_\text{TH}}=I\cdot R_2\$$, $$\R_{_\text{TH}}=R_1\$$

Your $$\I\$$ and $$\R_2\$$ is already a Norton source. The Norton source can be converted to its equivalent Thevenin source. Note that $$\R_1\$$ isn't needed to do that. This is entirely about $$\I\$$ and $$\R_2\$$:

simulate this circuit – Schematic created using CircuitLab

You are correct that $$\V_{_\text{TH}}=I\cdot R_2\$$. But incorrect when writing that $$\R_{_\text{TH}}=R_1\$$. Instead, $$\R_{_\text{TH}}=R_2\$$.

If you look, you should see that $$\R_1\$$ wasn't part of the conversion.

The goal is to have a single ideal voltage source and a single ideal resistance. So you need to add $$\R_1\$$ to $$\R_{_\text{TH}}\left[=R_2\right]\$$ together to get the final Thevenin equivalent resistance.

Let's call this value $$\R_{_\text{TH}}^{\:'}=R_{_\text{TH}}+R_1=R_2+R_1\$$.

You can convert this back to a Norton equivalent. The new ideal Norton current source will be $$\I_{_\text{N}}=\frac{V_{_\text{TH}}}{R_{_\text{TH}}^{\:'}}\$$. The Norton resistance will just be $$\R_{_\text{N}}=R_{_\text{TH}}^{\:'}\$$.